Functions Question - None

Question

Let

f(x) = \frac{x}{1-x}

and let

\alpha

be a real number. If

x_0 = \alpha

,

x_1 = f(x_0)

,

x_2 = f(x_1)

,

\ldots

and

x_{2011} = -\frac{1}{2012}

then the value of

\alpha

is

Accepted Answer

Step 1: Find the pattern by computing successive iterations.

Given f(x) = x/(1-x), let's compute the first few terms starting with x₀ = α.

x₁ = f(x₀) = f(α) = α/(1-α)

Step 2: Compute x₂ = f(x₁).

x₂ = f(x₁) = [α/(1-α)] / [1 - α/(1-α)]

= [α/(1-α)] / [(1-α-α)/(1-α)]

= [α/(1-α)] × [(1-α)/(1-2α)]

= α/(1-2α)

Step 3: Compute x₃ = f(x₂).

x₃ = f(x₂) = [α/(1-2α)] / [1 - α/(1-2α)]

= [α/(1-2α)] / [(1-2α-α)/(1-2α)]

= [α/(1-2α)] × [(1-2α)/(1-3α)]

= α/(1-3α)

Step 4: Compute x₄ = f(x₃).

x₄ = f(x₃) = [α/(1-3α)] / [1 - α/(1-3α)]

= α/(1-4α)

Step 5: Recognize the general pattern.

The pattern shows: xₙ = α/(1-nα)

Step 6: Apply the formula to x₂₀₁₁.

x₂₀₁₁ = α/(1-2011α)

Step 7: Use the given condition.

We're given that x₂₀₁₁ = -1/2012

Therefore: α/(1-2011α) = -1/2012

Step 8: Solve for α.

Cross-multiplying: 2012α = -(1-2011α)

2012α = -1 + 2011α

2012α - 2011α = -1

α = -1

Step 9: Verify the solution.

If α = -1: x₂₀₁₁ = -1/(1-2011(-1)) = -1/(1+2011) = -1/2012 ✓

∴ Answer: D

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